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Suppose that the average number of Facebook friends users have is normally distributed with a mean of 110 and a standard deviation of about 40. Assume forty-four in...

Question

Suppose that the average number of Facebook friends users have is normally distributed with a mean of 110 and a standard deviation of about 40. Assume forty-four individuals are randomly chosen. Answer the following questions. Round all answers to decimal places where possibleWhat is the distribution of *? € 110 6.0302 b. For the group of 44, find the probability that the average number of friends is more than 108.Find the third quartile for the average number of Facebook friends_ 114.0673 For

Suppose that the average number of Facebook friends users have is normally distributed with a mean of 110 and a standard deviation of about 40. Assume forty-four individuals are randomly chosen. Answer the following questions. Round all answers to decimal places where possible What is the distribution of *? € 110 6.0302 b. For the group of 44, find the probability that the average number of friends is more than 108. Find the third quartile for the average number of Facebook friends_ 114.0673 For part b), is the assumption that the distribution is normal necessary? Nol Yes



Answers

Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site. On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent.
a. Find the probability that the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning is at least 30.
b. Find the 95th percentile, and express it in a sentence.

Okay, So in this problem, we are testing a player's any sufficient evidence that, given the significant level, is pointing by the average baseball friend of 1% s square than 338. So we're testing at sample size of 50 high school student, and then we observe the sample Mean is 250. Even standard deviation is 43.2. So let's they don't know, hypothesis is not Ismaeel would be 338 and then the alternative hypotheses will be new greater than 338. So this is our claim. So this one is right tail test with a critical value at outfight you go to a 0.5, we're gonna look at the tea table over here. So we have Alba equal to playoff I, which is between these two value and then we get the d is negative 1.6 that I'm gonna take the right one, which is 1.65 So because the table is enough till table. So for the writing table, which is in our problem, I'm gonna take the ego to positive 1.65 So the critical value is positive 1.65 and then we have the test value ISS in this formula. So that's me, right? Right down, which is 300 feet. The miners 338 divided by the Houston at Aero 43.2. Oh verse Well, a 50. So I only computed it. Just approximately one play 96. Okay, so on test value is 1.96. And then he had ISS the significant level boy 05 at the value of 1.65 and then our test value of this one point. Nice eggs. So this region with on a project is not an A this region, we not me, Jake, It's not okay. So the decision is who gonna be Judge is not. And for the result is there is evidence that no Facebook friend, the Irish people friend off a person is Cueto them 338

Okay. So for this problem were given a density function as of tea and were asked to find the expectation of tea and standard deviation to start out. Well, look at what we need to do to find expectation. We're gonna integrate over our domain, which it's not written here, but as 13.4 up to 62. So we're gonna integrate from 13.4 to 62 this entire function, Uh, s of T times TV. Uh, DT. And so what? This is going to be since this is just a polynomial multiplying it by t is just gonna raise all of our powers by one. So we're gonna get this whole constant times five this one That's four times three attempts to our attempts t to the to t. Uh, and this last one will be 1 37.5 times tea, right? We're just multiplying t through this whole thing. Um, and we can move the constant part upside. So integrating this is a little bit annoying, but not too hard, Since all of the terms look like some form of a constant times are variable t to some power When we integrate this with respect to t We're just going to get see over plus one, do you, k plus one, right? So we just need toe up the power by one and divide the constant front by that. So to save us some space, I'm gonna go ahead and ah, just write out what this integral evaluates out to. So there's a lot of different things online that you can use to solve that or to check your work, But it would be just a little bit tedious to do manually. So our answer is gonna come out to be 31 0.741 and that's gonna be just our first part. So after we've got expectation, the next step in this problem is finding out, uh, the standard deviation. And so if we want standard deviation of X, what we really want is a squared off the variants, and we know that the variants is equal to expected t squared minus expected t squared. And so we already have expected t. It's this 31.7491 and we can easily square it. So what we need to do is find this expected t squared term expected t squared. And so what? That's gonna be very similar, Integral. It's going to be our density function S of T. But this time it's times t squared because we're trying to find expectation of T Square and again, all since this is a polynomial, all that we're gonna do is be using this formula again, right? And so to save a little bit of time, we'll just write down. But the results this from that we're gonna get 1141.5. And so this is not quite where we want to stop for the second part. This is just are expected t squared from that 1000. Ah, 141.5. We need to subtract off 31.74 91 squared. So we need to take This is 31.74 91 squared as that's that's are expected, t um, doing that will give us are variants. And so if you plug this into a calculator, you'll see that the variances that equal to 1 33.4946 and to get to the standard deviation then we want the square root of 1 33.4946 and so bad comes out to be oops, about 11.55 and so that will be our final answer for the second part of this question.

So in this question, were given a normally distributed population with men of 57.7 and a standard deviation, find the probability that a single randomly selected element X is less than 45. So this is a normally distributed population, so they can just convert that to our standard normal random variable, We have our population mean and standard definition. So that's probability oh z less than -1.052, which is .14 69 What? We were us to find the mean and standard deviation of the sample means that the mean of the sample mean is the population means Which is 57.7. The standard division is 12.1 over the square root of simple size, Which is 3.025. Find the probability that the mean of a sample of size 16 is less than 45, probability that the means of the sample is less than 45. So we take the sample means for for five minus 57 point 7/3 57.0.25 Which is equal to probably it easy, less than -4.19 that is zero.


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